1/* $NetBSD: qdivrem.c,v 1.2 2009/03/15 22:31:12 cegger Exp $ */ 2 3/*- 4 * Copyright (c) 1992, 1993 5 * The Regents of the University of California. All rights reserved. 6 * 7 * This software was developed by the Computer Systems Engineering group 8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 9 * contributed to Berkeley. 10 * 11 * Redistribution and use in source and binary forms, with or without 12 * modification, are permitted provided that the following conditions 13 * are met: 14 * 1. Redistributions of source code must retain the above copyright 15 * notice, this list of conditions and the following disclaimer. 16 * 2. Redistributions in binary form must reproduce the above copyright 17 * notice, this list of conditions and the following disclaimer in the 18 * documentation and/or other materials provided with the distribution. 19 * 3. Neither the name of the University nor the names of its contributors 20 * may be used to endorse or promote products derived from this software 21 * without specific prior written permission. 22 * 23 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 24 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 25 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 26 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 27 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 28 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 29 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 30 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 31 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 32 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 33 * SUCH DAMAGE. 34 */ 35 36#include <sys/cdefs.h> 37#if defined(LIBC_SCCS) && !defined(lint) 38#if 0 39static char sccsid[] = "@(#)qdivrem.c 8.1 (Berkeley) 6/4/93"; 40#else 41__RCSID("$NetBSD: qdivrem.c,v 1.2 2009/03/15 22:31:12 cegger Exp $"); 42#endif 43#endif /* LIBC_SCCS and not lint */ 44 45/* 46 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed), 47 * section 4.3.1, pp. 257--259. 48 */ 49 50#include "quad.h" 51 52#define B ((int)1 << HALF_BITS) /* digit base */ 53 54/* Combine two `digits' to make a single two-digit number. */ 55#define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b)) 56 57/* select a type for digits in base B: use unsigned short if they fit */ 58#if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff 59typedef unsigned short digit; 60#else 61typedef u_int digit; 62#endif 63 64static void shl __P((digit *p, int len, int sh)); 65 66/* 67 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v. 68 * 69 * We do this in base 2-sup-HALF_BITS, so that all intermediate products 70 * fit within u_int. As a consequence, the maximum length dividend and 71 * divisor are 4 `digits' in this base (they are shorter if they have 72 * leading zeros). 73 */ 74u_quad_t 75__qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq) 76{ 77 union uu tmp; 78 digit *u, *v, *q; 79 digit v1, v2; 80 u_int qhat, rhat, t; 81 int m, n, d, j, i; 82 digit uspace[5], vspace[5], qspace[5]; 83 84 /* 85 * Take care of special cases: divide by zero, and u < v. 86 */ 87 if (vq == 0) { 88 /* divide by zero. */ 89 static volatile const unsigned int zero = 0; 90 91 tmp.ul[H] = tmp.ul[L] = 1 / zero; 92 if (arq) 93 *arq = uq; 94 return (tmp.q); 95 } 96 if (uq < vq) { 97 if (arq) 98 *arq = uq; 99 return (0); 100 } 101 u = &uspace[0]; 102 v = &vspace[0]; 103 q = &qspace[0]; 104 105 /* 106 * Break dividend and divisor into digits in base B, then 107 * count leading zeros to determine m and n. When done, we 108 * will have: 109 * u = (u[1]u[2]...u[m+n]) sub B 110 * v = (v[1]v[2]...v[n]) sub B 111 * v[1] != 0 112 * 1 < n <= 4 (if n = 1, we use a different division algorithm) 113 * m >= 0 (otherwise u < v, which we already checked) 114 * m + n = 4 115 * and thus 116 * m = 4 - n <= 2 117 */ 118 tmp.uq = uq; 119 u[0] = 0; 120 u[1] = (digit)HHALF(tmp.ul[H]); 121 u[2] = (digit)LHALF(tmp.ul[H]); 122 u[3] = (digit)HHALF(tmp.ul[L]); 123 u[4] = (digit)LHALF(tmp.ul[L]); 124 tmp.uq = vq; 125 v[1] = (digit)HHALF(tmp.ul[H]); 126 v[2] = (digit)LHALF(tmp.ul[H]); 127 v[3] = (digit)HHALF(tmp.ul[L]); 128 v[4] = (digit)LHALF(tmp.ul[L]); 129 for (n = 4; v[1] == 0; v++) { 130 if (--n == 1) { 131 u_int rbj; /* r*B+u[j] (not root boy jim) */ 132 digit q1, q2, q3, q4; 133 134 /* 135 * Change of plan, per exercise 16. 136 * r = 0; 137 * for j = 1..4: 138 * q[j] = floor((r*B + u[j]) / v), 139 * r = (r*B + u[j]) % v; 140 * We unroll this completely here. 141 */ 142 t = v[2]; /* nonzero, by definition */ 143 q1 = (digit)(u[1] / t); 144 rbj = COMBINE(u[1] % t, u[2]); 145 q2 = (digit)(rbj / t); 146 rbj = COMBINE(rbj % t, u[3]); 147 q3 = (digit)(rbj / t); 148 rbj = COMBINE(rbj % t, u[4]); 149 q4 = (digit)(rbj / t); 150 if (arq) 151 *arq = rbj % t; 152 tmp.ul[H] = COMBINE(q1, q2); 153 tmp.ul[L] = COMBINE(q3, q4); 154 return (tmp.q); 155 } 156 } 157 158 /* 159 * By adjusting q once we determine m, we can guarantee that 160 * there is a complete four-digit quotient at &qspace[1] when 161 * we finally stop. 162 */ 163 for (m = 4 - n; u[1] == 0; u++) 164 m--; 165 for (i = 4 - m; --i >= 0;) 166 q[i] = 0; 167 q += 4 - m; 168 169 /* 170 * Here we run Program D, translated from MIX to C and acquiring 171 * a few minor changes. 172 * 173 * D1: choose multiplier 1 << d to ensure v[1] >= B/2. 174 */ 175 d = 0; 176 for (t = v[1]; t < B / 2; t <<= 1) 177 d++; 178 if (d > 0) { 179 shl(&u[0], m + n, d); /* u <<= d */ 180 shl(&v[1], n - 1, d); /* v <<= d */ 181 } 182 /* 183 * D2: j = 0. 184 */ 185 j = 0; 186 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ 187 v2 = v[2]; /* for D3 */ 188 do { 189 digit uj0, uj1, uj2; 190 191 /* 192 * D3: Calculate qhat (\^q, in TeX notation). 193 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and 194 * let rhat = (u[j]*B + u[j+1]) mod v[1]. 195 * While rhat < B and v[2]*qhat > rhat*B+u[j+2], 196 * decrement qhat and increase rhat correspondingly. 197 * Note that if rhat >= B, v[2]*qhat < rhat*B. 198 */ 199 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */ 200 uj1 = u[j + 1]; /* for D3 only */ 201 uj2 = u[j + 2]; /* for D3 only */ 202 if (uj0 == v1) { 203 qhat = B; 204 rhat = uj1; 205 goto qhat_too_big; 206 } else { 207 u_int nn = COMBINE(uj0, uj1); 208 qhat = nn / v1; 209 rhat = nn % v1; 210 } 211 while (v2 * qhat > COMBINE(rhat, uj2)) { 212 qhat_too_big: 213 qhat--; 214 if ((rhat += v1) >= B) 215 break; 216 } 217 /* 218 * D4: Multiply and subtract. 219 * The variable `t' holds any borrows across the loop. 220 * We split this up so that we do not require v[0] = 0, 221 * and to eliminate a final special case. 222 */ 223 for (t = 0, i = n; i > 0; i--) { 224 t = u[i + j] - v[i] * qhat - t; 225 u[i + j] = (digit)LHALF(t); 226 t = (B - HHALF(t)) & (B - 1); 227 } 228 t = u[j] - t; 229 u[j] = (digit)LHALF(t); 230 /* 231 * D5: test remainder. 232 * There is a borrow if and only if HHALF(t) is nonzero; 233 * in that (rare) case, qhat was too large (by exactly 1). 234 * Fix it by adding v[1..n] to u[j..j+n]. 235 */ 236 if (HHALF(t)) { 237 qhat--; 238 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ 239 t += u[i + j] + v[i]; 240 u[i + j] = (digit)LHALF(t); 241 t = HHALF(t); 242 } 243 u[j] = (digit)LHALF(u[j] + t); 244 } 245 q[j] = (digit)qhat; 246 } while (++j <= m); /* D7: loop on j. */ 247 248 /* 249 * If caller wants the remainder, we have to calculate it as 250 * u[m..m+n] >> d (this is at most n digits and thus fits in 251 * u[m+1..m+n], but we may need more source digits). 252 */ 253 if (arq) { 254 if (d) { 255 for (i = m + n; i > m; --i) 256 u[i] = (digit)(((u_int)u[i] >> d) | 257 LHALF((u_int)u[i - 1] << (HALF_BITS - d))); 258 u[i] = 0; 259 } 260 tmp.ul[H] = COMBINE(uspace[1], uspace[2]); 261 tmp.ul[L] = COMBINE(uspace[3], uspace[4]); 262 *arq = tmp.q; 263 } 264 265 tmp.ul[H] = COMBINE(qspace[1], qspace[2]); 266 tmp.ul[L] = COMBINE(qspace[3], qspace[4]); 267 return (tmp.q); 268} 269 270/* 271 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that 272 * `fall out' the left (there never will be any such anyway). 273 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS. 274 */ 275static void 276shl(digit *p, int len, int sh) 277{ 278 int i; 279 280 for (i = 0; i < len; i++) 281 p[i] = (digit)(LHALF((u_int)p[i] << sh) | 282 ((u_int)p[i + 1] >> (HALF_BITS - sh))); 283 p[i] = (digit)(LHALF((u_int)p[i] << sh)); 284} 285